For each n ≥ 1 define f_n : ℝ → ℝ by \(\rm f_n(x)=\frac{x^2}{√{x^2+\frac{1}{n}}}, \) x ∈ ℝ

where √ denotes the non-negative square root. Wherever \(\rm \lim_{n \rightarrow \infty}f_n(x)\) exists, denote it by f(x). Which of the following statements is true? 

Concept:

Limit of a Sequence of Functions:

1. Let \(\{f_n\}\) be a sequence of functions defined on a set D . We say that \(f_n\) converges pointwise to a function

 f  on D if, for every x \(\in\) D

\(\lim_{n \to \infty} f_n(x) = f(x).\)

2. A stronger form of convergence is uniform convergence. The sequence \(\{f_n\}\) converges uniformly to a function  f  on D if

\(\lim_{n \to \infty} \sup_{x \in D} |f_n(x) - f(x)| = 0.\)

Explanation: The problem gives a sequence of functions\( f_n: \mathbb{R} \to \mathbb{R}\) defined by

\(f_n(x) = \frac{x^2}{\sqrt{x^2 + \frac{1}{n}}}\)

and asks about the limit of \(f_n(x) \) as \(n \to \infty \), denoted by \( f(x)\) . We are tasked with determining which statement about f(x) is true.

We are asked to take the limit \(n \to \infty \) of the function:

 \(f_n(x) = \frac{x^2}{\sqrt{x^2 + \frac{1}{n}}} \)   

As \(n \to \infty \), the term \(\frac{1}{n} \to 0 \). So, for large n , the function \(f_n(x) \) approaches

 \(lim_{n \to \infty} f_n(x) = \frac{x^2}{\sqrt{x^2}} = \frac{x^2}{|x|}\)
 

Case 1: \(x \neq 0\)

For \(x \neq 0\) we have, \( f(x) = \frac{x^2}{|x|} = |x|\)
 

Case 2: \(x =0\)
When \(x =0\) , the function becomes \(f_n(0) = \frac{0^2}{\sqrt{0^2 + \frac{1}{n}}} = 0\)

Therefore, as \(n \to \infty \), we get f(0) = 0 .

The function f(x) , \(n \to \infty \) , is given by

   
  \( f(x) = \begin{cases} |x|, & \text{if } x \neq 0 \\ 0, & \text{if } x = 0 \end{cases} \)

This function is equal to |x| for all \(x \in \mathbb{R} \), 

Therefore, The correct option is 4).